Question

The vertices of a triangle $OBC$ are $O(0,0),B(-3,-1),C(-1,-3)$. Find the equation of the line parallel to $BC$ and intersecting the sides $OB$ and $OC$, whose perpendicular distance from the point $(0,0)$ is half.

Answer 1

Equation of the line BC is given by

$y-(-1)=\frac{-3-(-1)}{-1-(-3)}(x+3)$

$\Rightarrow y+1=\frac{-2}{2}(x+3)$

$\Rightarrow y+1+x+3=0$

$\Rightarrow x+y+4=\mathrm{0........}\left(1\right)$

Any line parallel to BC (say B'C') is given by

$x+y+\lambda =\mathrm{0..........}\left(2\right)$

$\therefore$ Length of the perpendicular from (0,0) on (2) is

$\frac{\lambda }{\sqrt{2}}=\pm \frac{1}{2}$

$\Rightarrow \lambda =\pm \frac{1}{\sqrt{2}}=\pm \frac{\sqrt{2}}{2}$

Choosing + sign as - sign would mean line on the other side of O i.e, not intersecting OB and OC.

Hence the equation of B'C' is given by

$x+y+\frac{\sqrt{2}}{2}=0$

$2x+2y+\sqrt{2}=0$